l parucular brand of television set, the distribution of failures has a mean ont
ID: 3329490 • Letter: L
Question
l parucular brand of television set, the distribution of failures has a mean onths with a standard deviation of 12 months. With t to this distribution, 140 months corresponds to what finish in the top 20% 64. Weig ht lifting. Ina tion of total weight a standard deviation weight lifted for a co the competition? z-score? Product reliability. Redo Exercise 51 using 130 months instead of 140 Distribution of heights. 65. Strength of cable Assume that in the NBA the distri- breaking point of 1 8 pounds. What we 95% of the cables 53. tion of heights has a mean of 6 feet, 8 inches, and a standard deviation of 3 inches. If there are 324 players in the league, how many players do you expect to be over 7 feet tall? 54. Distribution of heights. Redo Exercise 53, but now we want to know how many players you would expect to be over 6 feet, 6 inches. 55. Distribution of heart rates. Assume that the distribu- tion of 21-year-old women's heart rates at rest is 68 beats per minute with a standard deviation of 4 beats per minute. If 200 women are examined, how many would you expect to have a heart rate of less than 70? 56, , Distribution of heart rates. Redo Exercise 55, but now k how many you would expect to have a heart rate less 66. Strength of c 180 pounds, the expect 90% of th than 75. lanath of a human 67 Rainieet citvExplanation / Answer
53.
the PDF of is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard is a with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 80
standard Deviation ( sd )= 3
P(X > 84) = (84-80)/3
= 4/3 = 1.3333
= P ( Z >1.3333) From Standard Normal Table
= 0.0912
54.
P(X > 78) = (78-80)/3
= -2/3 = -0.6667
= P ( Z >-0.6667) From Standard Normal Table
= 0.7475
55.
the PDF of is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard is a with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 68
standard Deviation ( sd )= 4
P(X < 70) = (70-68)/4
= 2/4= 0.5
= P ( Z <0.5) From Standard Normal Table
= 0.6915
56.
P(X < 75) = (75-68)/4
= 7/4= 1.75
= P ( Z <1.75) From Standard Normal Table
= 0.9599
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